x=1,y=0,z=1;
x=133/7,y=-24/7,z=97/7
x+2z=3 x+y+z=2 2x+y=2 3x-y-4z=5 2y+z=7 2x+3y-2z=0
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相关问题
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三元一次方程求解：①x+2z=3,2x+y=2,2y+z=7.②x+y+z=2,3x-y-4z=5,2x+3y-2z=0
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x-5y-3z=0 y+z-3x=3 x+y+3z=14 y+4z=3 z+x-3y=5 2x+y+z=7 2x-z=1
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证明：（y+z-2x）3+（z+x-2y）3+（x+y-2z）3=3（y+z-2x）（z+x-2y）（x+y-2z）．
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2x-y=7 x+y+z=25x+3y+z=2 x-2y+z=-1 3x-4z=4 x+2y+3z=-1x+y-z=6x
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x+y+z=2 ① 2x+3y+z=38 ① x+y=3 3x-y-4z=5 ② 3x+4y+2z+56 ② y+z=5
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x+y-z=0 2x-3y+2z=5 x+2y-z=3
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x+y-2z=0 3x-2y-2z=-5 2x+y-z=3
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{x+2y+4z=0{2x-y+3z=-10{3x+y+2z=5
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2x-3y+4z=2 4x+y+7z=3 8x+2y+3z=-5
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(1)2X+4Y+3Z=93X-2Y+5Z=115X-6Y+7Z=13(2)X:Y=3:2Y:Z=5:4X+Y+Z=66